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Proof step 2 from proof byte-mult-pos (previous step)

0 ≤ i, 0 ≤ j ⊦ i2b(i * 256 + j) = i2b(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i2b(i) = i2b(j) ↔ b2i(i2b(i)) = b2i(i2b(j))
Proof step 4:
0 ≤ i, 0 ≤ j ⊦ b2i(i2b(i * 256 + j)) = b2i(i2b(j))

Rule argument

name: i2b-equal rotate: F subst: [j, i] → [j, i * 256 + j]

Additional information

Used terms: j, i * 256 + j